# How do you use a Taylor series to prove Euler's formula?

Nov 15, 2014

Euler's Formula

${e}^{i \theta} = \cos \theta + i \sin \theta$

Let us first review some useful power series.

e^x=1/{0!}+x/{1!}+x^2/{2!}+cdots

cos x=1/{0!}-x^2/{2!}+x^4/{4!}-cdots

sin x=x/{1!}-x^3/{3!}+x^5/{5!}-cdots

Now, we are ready to prove Euler's Formula.

Proof

By rewriting as a power series,

e^{i theta}=1/{0!}+(i theta)/{1!}+(itheta)^2/{2!}+(i theta)^3/{3!}+(i theta)^4/{4!}+(i theta)^5/{5!}+cdots

by distributing the powers,

=1/{0!}+i theta/{1!}+i^2 theta^2/{2!}+i^3 theta^3/{3!}+i^4 theta^4/{4!}+i^5 theta^5/{5!}+cdots

by ${i}^{2} = - 1$

=1/{0!}+i theta/{1!}-theta^2/{2!}-i theta^3/{3!}+theta^4/{4!}+i theta^5/{5!}-cdots

by separating the real part and the imaginary part,

=(1/{0!}-theta^2/{2!}+theta^4/{4!}-cdots)+i(theta/{1!}-theta^3/{3!}+theta^5/{5!}-cdots)

by identifying the power series,

$= \cos \theta + i \sin \theta$

Hence, we have Euler's Formula

${e}^{i \theta} = \cos \theta + i \sin \theta$.

I hope that this was helpful.